Lambda -> n + pi^0 decay

[BarakSh]

The \Lambda baryon (quark content uds) decays into n + \pi^0 or p + \pi^-. In the case \Lambda \to p + \pi^-, the s quark decays into a u quark, releasing a W^- in the process (which subsequently decays into a \pi^- meson). What happens in the \Lambda \to n + \pi^0 case? (I tried Google, but couldn't find anything about this specific decay.)

Thanks!

[phyzguy]

At the quark level, these two decays are the same. The s quark decays into a u quark plus a W-, which decays into and up-bar antiquark and a down quark. In the first case, the u quark from the s decay ends up with the u and d from the original lambda, and the upbar and down pair together, so we end up with a proton and a pi-. In the second case, the d quark from the W- decay ends up with the u and d from the original lambda, and the upbar and up quark pair together, so we end up with a neutron and a pi-0.